Dif. Eq. General solution is sum of homogeneous and particular solutions Why is the general solution to ordinary first order differential equations a sum of homogeneous(Setting the inhomogeneous term to 0) and particular(satisfies the differential equation but not necessarily the initial conditions) solutions?
$x=x_{h}+x_{p}$
I have tried finding a proof on the Internet but without a result.
Can someone give me a not too complicated mathematical proof(If possible) to why this relation is valid?
Thank you!
 A: Without initial or boundary conditions, the solution to your average DE is a family of solutions. For example, the solution set to $x' = x$ is $\{Ae^t\}$, a set parametrised by the value $A$.
So, if $x_p$ is a particular solution to a given linear DE, and $x_h$ is a generic element of the solution set of the homogeneous DE, then at the very least you can show that $x = x_h + x_p$ represents a solution set to the original DE, because $x' = x_h' + x_p'$ and so on for all further derivatives. In other words, if the original DE is $a_0 x + a_1 x' + a_2 x'' + \ldots + a_n x^{(n)} = c$, then we have:
$\begin{eqnarray}a_0 x + a_1 x' + a_2 x'' + \ldots + a_n x^{(n)} & = & a_0 (x_h + x_p) + a_1 (x_h + x_p)' + \ldots + a_n (x_h + x_p)^{(n)} \\
 & = & a_0 x_h + a_0 x_p + a_1 x_h' + a_1 x_p' + \ldots + a_n x_h^{(n)} + a_n x_p^{(n)} \\
& = & \left( a_0 x_h + a_1 x_h' + \ldots + a_n x_h^{(n)} \right) + \left( a_0 x_p + a_1 x_p' + \ldots + a_n x_p^{(n)} \right) \\
& = & 0 + C\end{eqnarray}$
Because $x_h$ is a solution to the homogeneous DE $a_0 x_h + a_1 x_h' + \ldots + a_n x_h^{(n)} = 0$ and $x_p$ is a solution to the linear DE.
Thus, we at least know that if we add a particular solution and a homogeneous solution together, we get another solution to the DE. So if we add the entire homogeneous solution set to the particular solution we generate a set of functions that are also solutions to the original DE. So as long as we can then pick an element of that set that satisfies our boundary/initial conditions, we can generate the required solution to the DE.
Proof that (a) the set of all $x_h + x_p$ is the full solution set to the DE, and (b) that the method generates unique solutions are slightly trickier exercises.
A: Let us denote the linear differential operator by $D$. We write its corresponding differential equation as $D(x)=b$, with full solution set $\Omega$.
Let $p$ be any particular solution to the equation $D(x)=b$, such that $D(p)=b$.
Now consider the set $ S=\{p+h|h    \text{ satisfies the homogeneous equation}\}$. We will show that this is equal to the solution set. We will show set inclusion both ways.
First we will show that if a general solution exists, it is in this set.
Assume that $s\in \Omega$, the full solution set to the differential equation, then consider:
$D(s-p)=D(s)-D(p)=b-b=0$.
This means that $s-p$ is a homogeneous solution, therefore we know $s=p+(s-p)\in S$. Hence $ \Omega \subseteq S$
Finally, by linearity of differentiation we have for any $p+h \in S$ :
$D(p+h)=D(p)+D(h)=b+0=b$.
So $S\subseteq \Omega$.
Thus by mutual inclusion $S=\Omega$ $\square$.
